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[parent] examples of Einstein summation notation (Example)

Some examples of applying the Einstein summation notation.

Example 1. Let us consider the quantity

         α  β
S =  aαβx  x

for a three dimensional space. Since the index α occurs as both a subscript and a superscript, we sum on α from 1 to 3. This yields

S =  a1βx1xβ + a2βx2x β + a3βx3x β

Now each term of S is such that β is both a subscript and superscript. Summing on β from 1 to 3 as prescribed by our summation convention yields the quadratic form

S = a11x1x1 + a12x1x2 + a13x1x3
         2 1       2 2       2 3
   +a21x x  + a22x x  + a23x x
   +a31x3x1 + a32x3x2 + a33x3x3

Example 2. If x1,x2,x3,…,xn is a set of independent variables, then

∂x1-   ∂x2-   ∂x3-         ∂xn-
∂x1  = ∂x2  = ∂x3  = ⋅⋅⋅ = ∂xn  = 1

and if i≠j

∂x1-          ∂xi-
∂x2 =  0,     ∂xj = 0

We may write

         {
∂xi     i  =   1     if  i = j
---j ≡ δj
∂x         =   0     if  i ⁄= j
(1)

The symbol δji is called the Kronecker delta. We have

δαα =  δ11 + δ22 + ⋅⋅⋅ + δnn = n

Let us now assume that the quadratic form at the end of example 1 vanishes identically for all values of the independent variables x1,x2, x3, and a ij to be constant. Differentiating S = aαβxαxβ = 0 with respect to a given variable, say xi, yields

               β            α
-∂S-=  aαβxα ∂x--+ a αβxβ∂x-- = 0
∂xi          ∂xi         ∂xi
    ∂S--       α β        β α
    ∂xi =  aαβx δi + aα βx δi = 0
          ∂S
          --i-= aαixα + aiβxβ = 0
          ∂x

Now differentiating with respect to xi yields

--∂2S--       α       β
∂xj ∂xi = aαiδj + aiβδi = 0

so that aji + aij = 0 or aij = aji for i,j = 1, 2, 3.

Example 3. We define 𝜖ij,i,j = 1, 2, to have the following numerical values: Let 𝜖11 = 𝜖22 = 0,𝜖12 = 1,𝜖21 = 1. We now consider the expression

D =  𝜖ija1ia2j
(2)

Expanding (2) by use of our summation convention yields

D  = 𝜖11a1a2 + 𝜖12a1 a2+ 𝜖21a1a2 + 𝜖22a1a2=  a1a2−  a1a2
         1 1       1 2      2 1       2 2    1 2    2 1

The reader who is familiar with second-order determinants quickly recognizes that

          |       |
          |a1  a1 |
𝜖ija1ia2j = || 12   22 ||
           a1  a2
(3)

Example 4. The system of equations

(                            )
   y1  =   y1(x1,x2,⋅⋅⋅ ,xn)
||  y2  =   y2(x1,x2,⋅⋅⋅ ,xn) ||
|   ..   ..          ..         |
(   .   .          .         )
   yn  =   yn(x1,x2,⋅⋅⋅ ,xn)
(4)

represents a coordinate transformation from an (x1,x2,⋅⋅⋅,xn) coordinate system to a (y1,y2,⋅⋅⋅,yn) coordinate system. From the calculus we have

  i   ∂yi   1   ∂yi   2         ∂yi   n
dy  = ---1dx  + ---2dx  + ⋅⋅⋅ + --n-dx      i = 1,2, ⋅⋅⋅ ,n
      ∂x        ∂x              ∂x

         i
  i   ∂y--  α
dy =  ∂yα dx

The α in the term   i
∂∂yxα is to be considered as a subscript. If, furthermore, the xi, i = 1, 2,⋅⋅⋅,n, can be solved for the y1,y2,⋅⋅⋅,yn, and assuming differentiability of the xi with respect to each yi, one obtains

∂yi         ∂yi ∂xα
----≡ δij =  --------
∂yj         ∂xα ∂yj

Differentiating this expression with respect to yk yields

    ∂yi--∂2xα--   --∂2yi--∂xβ-∂xα-
0 = ∂xα ∂yk∂yi +  ∂xβ∂x α ∂yk ∂yj

Multiplying both sides of this equation by ∂xσi
 ∂y amd summing on the inex i yields

     ∂xσ ∂yi  ∂2xα       ∂2yi  ∂xβ ∂xα ∂x σ
0 =  --i---α---k---j + ---β--α---k----j---i
     ∂y  ∂x  ∂y ∂y     ∂x  ∂x  ∂y  ∂y  ∂y

or

        ∂2xα       ∂2yi  ∂xβ ∂xα ∂x σ
0 = δασ-------+  --------------------
       ∂yk∂yj    ∂xβ ∂xα ∂yk ∂yj ∂yi

which yields

 ∂2xσ        ∂2yi   ∂xβ ∂xα ∂xσ
-------=  − --------------------
∂yk∂yj      ∂xβ∂x α ∂yk ∂yj ∂yi

In particular, if y = f(x), then

 2        2 (    )3
d-x-= − -d-y  dx-
dy2     dx2   dy

0.1 References

[1] Lass, Harry. ”Elements of pure and applied mathematics” New York: McGraw-Hill Companies, 1957.

This entry is a derivative of the Public domain work [1].


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